Problem: $f(x,y) = \sin^2(x) - \cos(y)$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sin(2x)$ (Choice B) B $0$ (Choice C) C $\sin(x + y)$ (Choice D) D $\sin(y)$
Solution: We want to find $\dfrac{\partial f}{\partial y}$, which is the partial derivative of $f$ with respect to $y$. When we take a partial derivative with respect to $y$, we treat $x$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial y} \left[ \sin^2(x) \right] = 0 \\ \\ &\dfrac{\partial}{\partial y} \left[ -\cos(y) \right] = \sin(y) \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial y} = 0 + \sin(y) = \sin(y)$